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Physics M.C.Q [1M]

MODEL PAPER 2 · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 12 questions.

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M.C.Q [1M]

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12 questions · timed · auto-graded

MCQ 11 Mark
Point charges $+4 q,-q$ and $+4 q$ are kept on the $X$-axis at points $x=0, x=a$ and $x=2$ a respectively:
  • A
    all the charges are in unstable equilibrium
  • B
    all the charges are in stable equilibrium
  • C
    none of the charges is in equilibrium
  • D
    only - q is in stable equilibrium
Answer
(a) all the charges are in unstable equilibrium
Explanation: The net force on each charge is zero. Therefore, all the charges are in equilibrium. If we slightly displace the charge -q to the right, the net force of attraction will further displace it to the right i.e., away from its mean positive. The equilibrium is, therefore, unstable.
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MCQ 21 Mark
An astronomical telescope of ten fold angular magnification has a length of 44 cm. The focal length of the objective is
  • A
    44cm
  • B
    440cm
  • C
    4cm
  • D
    40cm
Answer
(d) 40cm
Explanation:
\[
L=f_o+f_e=44 \text { and }|m|=\frac{f_o}{f_e}=10
\] This gives $f_o=40 cm$
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MCQ 31 Mark
According to Joule's law, if potential difference across a conductor of material of resistivity remains constant, then heat produced in the conductor is directly proportional to
  • A
    $\rho$
  • B
    $\frac{1}{\sqrt{\rho}}$
  • C
    $\rho^{-1}$
  • D
    $\rho^2$
Answer
(c) $\rho^{-1}$
Explanation: $\rho^{-1}$
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MCQ 41 Mark
Time period of oscillation of a magnetic needle is
  • A
    $T=\sqrt{\frac{I}{M B}}$
  • B
    $T=\pi \sqrt{\frac{M B}{I}}$
  • C
    $T=2 \pi \sqrt{\frac{M B}{I}}$
  • D
    $T=2 \pi \sqrt{\frac{I}{M B}}$
Answer
(d) $T=2 \pi \sqrt{\frac{I}{M B}}$
Explanation: Time period of oscillation of a magnetic needle is $T=2 \pi \sqrt{\frac{I}{M B}}$
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MCQ 51 Mark
C and Si both have the same lattice structure, having 4 bonding electrons in each. However, C is an insulator whereas Si is an intrinsic semiconductor. This is because
A. In case of C the valence band is not completely filled at absolute zero temperature.
B. In case of C the conduction band is partly filled even at absolute zero temperature.
C. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.
D. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.
  • A
    Option B
  • B
    Option A
  • C
    Option D
  • D
    Option C
Answer
(d) Option C
Explanation: ${ }^6 C : 1 s^2 2 s^2 2 p^2$
${ }^{14} Si : 1 s^2 2 s^2 2 p^6 3 s^2 3 p^2$
The energy required to take out an electron from the 3rd orbit of Si is much smaller than to take out an electron from the 2nd orbit of C. So, Si has a significant number of free electrons while C has a negligibly small number of free electrons.

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MCQ 61 Mark
The electrostatic force between the metal plates of an isolated parallel capacitor C having a charge Q and area A is:
  • A
    independent of the distance between the plates
  • B
    inversely proportional to the distance between the plates
  • C
    proportional to the square root of the distance between the plates
  • D
    linearly proportional to the distance between the plates
Answer
(a) independent of the distance between the plates
Explanation: $F=Q E=Q \frac{\sigma}{2 \varepsilon_0}=Q \frac{Q}{2 A \varepsilon_0}=\frac{Q^2}{2 A \varepsilon_0}$
F is independent of the distance between the plates.
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MCQ 71 Mark
Consider sunlight incident on a slit of width $10^4$ A. The image seen through the slit shall
  • A
    only be a diffused slit white in colour
  • B
    a bright slit white at the center diffusing to regions of different colours
  • C
    a bright slit white at the center diffusing to zero intensities at the edges
  • D
    be a fine sharp slit white in colour at the center
Answer
(d) be a fine sharp slit white in colour at the center
Explanation: Width of slit $10^4 \stackrel{o}{A}=10,000 \stackrel{o}{A}$
Wavelength of visible light varies from 4000 to 8000$\stackrel{o}{A}$ . As the width of slit 10000 $\stackrel{o}{A}$ is comparable to that of wavelength of visible light i.e. 8000 $\stackrel{o}{A}$ . Hence the diffraction occurs with maxima at the centre. So at the centre all colours appear ,mixing of colors forms white colour at the centre.
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MCQ 81 Mark
The susceptibility of a paramagnetic material is $\chi$ at $27^{\circ} C$. At what temperature will its susceptibility be $\frac{\chi}{2}$ ?
  • A
    $54^{\circ} C$
  • B
    $327^{\circ} C$
  • C
    $237^{\circ} C$
  • D
    $1600^{\circ} C$
Answer
(b) $327^{\circ} C$
Explanation: $\frac{\chi_2}{\chi_1}=\frac{T_2}{T_1}$
$T _2=\frac{\chi_1}{\chi_2} \cdot T_1=\frac{\chi}{\chi / 2}(273+27) K =600 K=327^{\circ} C$
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MCQ 91 Mark
Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation $I = I _0 \sin$ $\Omega t$, where $I _0=10 A$ and $\omega=100 \pi rad s ^{-1}$. The maximum value of emf in the second coil is
  • A
    $12 \pi$
  • B
    $2 \pi$
  • C
    $5 \pi$
  • D
    $6 \pi$
Answer
(c) $5 \pi$
Explanation: $\varepsilon=M \frac{d I}{d t}$
$\begin{array}{l}=M \frac{d}{d t}\left[I_0 \sin \omega t\right]=M I_0 \omega \cos \omega t \\ E _{\max }=M I_0 \omega[\text { Max. value of } \cos \omega t=1] \\ =0.005 \times 10 \times 100 \pi=5 \pi V\end{array}$
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MCQ 101 Mark
A galvanometer of resistance 25Ω galvanometer is given by a 2.5Ω wire. The part of total current $I _0$ that flows through the galvanometer is given by
  • A
    $\frac{I}{I_0}=\frac{2}{11}$
  • B
    $\frac{I}{I_0}=\frac{4}{11}$
  • C
    $\frac{I}{I_0}=\frac{1}{11}$
  • D
    $\frac{I}{I_0}=\frac{3}{11}$
Answer
(c) $\frac{I}{I_0}=\frac{1}{11}$
Explanation:
Image

$\begin{array}{l}I=\frac{I_o \times 2.5}{(25+2.5)}=I_o \times \frac{25}{275}=\frac{1}{11} \times I_o \\ \Rightarrow \frac{ I }{ I _0}=\frac{1}{11}\end{array}$
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MCQ 111 Mark
Which of the following principle is used in optical fibre?
  • A
    Total internal reflection
  • B
    Scattering
  • C
    Interference
  • D
    Diffraction
Answer
(a) Total internal reflection
Explanation: Total internal reflection principle is used in optical fibre.
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MCQ 121 Mark
The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R _1$ will be
Image
  • A
    2.5 A
  • B
    10.0 A
  • C
    1.43 A
  • D
    3.13 A
Answer
(a) 2.5 A
Explanation: $D_1$ is reverse biased and $D_2$ is forward biased. $D_1$ blocks current. Hence, Current will flow through 10 V cell, $R _1, D _2$ and $R _3$.
$\begin{array}{l}\therefore I =\frac{\varepsilon}{R_1+R_3} \\ =\frac{10 V}{(2+2) \Omega}=2.5 A\end{array}$
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